therefore I understand that if you role a typical pair of dice, your possibilities of obtaining Snake eyes (double 1s) is $1$ in $36$. What I"m not certain of is how to carry out the to figure out your opportunities of rolling line Eyes at least once throughout a collection of rolls. I understand if I roll the dice $36$ time it won"t cause a $100\%$ possibility of rolling snake Eyes, and also while i imagine it"s in the top nineties, I"d favor to figure out exactly how i can not qualify it is.

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The probability the hitting that at the very least once is $1$ minus the probabilty that never hitting it.

Every time you roll the dice, you have actually a $35/36$ possibility of not hitting it. If you role the dice $n$ times, climate the only case where you have actually never hit it, is when you have not hit that every single time.

The probabilty of not hitting v $2$ roll is for this reason $35/36 imes 35/36$, the probabilty of no hitting with $3$ rolfes is $35/36 imes 35/36 imes 35/36=(35/36)^3$ and also so ~ above till $(35/36)^n$.

Thus the probability that hitting it at least once is $1-(35/36)^n$ where $n$ is the number of throws.

After $164$ throws, the probability of hitting that at the very least once is $99\%$

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edited Oct 4 "18 at 13:00

J. M. Ain't a civicpride-kusatsu.netematician
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reply Oct 3 "18 at 13:19

b00n heTb00n heT
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The other answers explain the general formula because that the probability of never ever rolling snake eyes in a series of $n$ rolls.

However, you likewise ask specifically about the instance $n=36$, i.e. If you have actually a $1$ in $k$ chance of success, what is your opportunity of obtaining at least one success in $k$ trials? It transforms out the the answer come this inquiry is quite similar for any kind of reasonably huge value that $k$.

It is $1-ig(1-frac1kig)^k$, and $ig(1-frac1kig)^k$ converges come $e^-1$. Therefore the probability will be about $1-e^-1approx 63.2\%$, and this approximation will certainly get better the bigger $k$ is. (For $k=36$ the actual answer is $63.7\%$.)

answer Oct 3 "18 at 13:40

especially LimeEspecially Lime
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If you roll $n$ times, climate the probability of rolling line eyes at least once is $1-left(frac3536 ight)^n$, together you either roll snake eye at the very least once or not at all (so the probability of this two events should sum to $1$), and also the probability of never rolling line eyes is the same as requiring that you roll among the various other $35$ possible outcomes on every roll.

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reply Oct 3 "18 in ~ 13:22
Sam StreeterSam Streeter
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